I want to declare type definition for a member function signature. Global function typedefs look like this:
typedef int (function_signature)(int, int);
typedef int (*function_pointer) (int, int);
But I'm not able to the same thing for a member function:
typedef int (foo::memberf_signature)(int, int); // memberf_pointer is not a member of foo
typedef int (foo::*memberf_pointer)(int, int);
It sounds logically to me, because foo:: is the syntax to access a member in the class foo.
How can I typedef just the signature?
For questions regarding the awkward function pointer syntax, I personally use a cheat-sheet: The Function Pointers Tutorial (downloadable here, thanks to Vector for pointing it out).
The signature of a member function, however, is a bit different from the signature of a regular function, as you experienced.
As you probably know, a member function has a hidden parameter, this, whose type need be specified.
// C++11 and above.
using Member = int (Foo::*)(int, int);
// C++03 and below.
typedef int (Foo::*Member)(int, int);
does let you specify that the first element passed to the function will be a Foo* (and thus your method really takes 3 arguments, when you think of it, not just 2.
However there is another reason too, for forcing you to specify the type.
A function pointer might refer to a virtual function, in which case things can get quite complicated. Therefore, the very size of the in-memory representation changes depending on the type of function. Indeed, on Visual Studio, a function pointer's size might vary between 1 and 4 times the size of a regular pointer. This depends on whether the function is virtual, notably.
Therefore, the class the function refers to is part of the signature, and there is no work-around.